Question

What should be the duration of a year if the distance between earth and the sun gets doubled the present distance?

Answer 1

Hint Kepler gave a relation between the time period of an object moving in a circle of radius R and its radius; use that to determine the change in time period of the earth.

Complete step-by-step solution
When an object moves around another object in a circle of radius r, a force act on it
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
This force is equal to the centripetal force acting on an object,
$\dfrac{{m{v^2}}}{r} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Solving the above equation, we get
${\omega ^2}{r^2} = \dfrac{{G{m_1}}}{r}$
${(\dfrac{{2\pi }}{T})^2}{r^2} = \dfrac{{G{m_1}}}{r}$

$$\dfrac{{4{\pi ^2}}}{{{T^2}}}{r^2} = \dfrac{{G{m_1}}}{r} \\\ {T^2} = k{R^3} \\\$$

This is known as the Kepler’s third Law of Planetary motion.
When the radius between the earth and the sun gets doubled, its time period will also increase as,

$${T_{new}} = k{(2R)^{\dfrac{3}{2}}} \\\ {T_{new}} = \sqrt 8 T \\\ {T_{new}} = 2.82T \\\$$

Therefore, the new duration of year will be 2.82 times the original year i.e. 1032 days per year.

Note When an object moves away from its axis of rotation, its angular momentum is conserved, but its moment of inertia increases. This will result in decreased angular velocity of the earth and hence it will take more time to complete 1 year